Clearly, one option is that $\psi(\xi_1,\dots,\xi_n) = \xi_1^{p_1} \cdots \xi_n^{p_n}$
such that $p_1+\cdots+p_n = c$, but can one exclude any other form?
How do one take advantage of the fact that $\psi$ is algebraic in the $\xi_i$?

2 Answers
2

So, here is my stab at a proof, which actually do not require algebraicness of $\psi$

Notice that we have $|\psi(\xi_1, \dots \xi_n)| = 1$ whenever $\xi_1 \cdots \xi_n = 1.$
Thus, using the homogeneity property, we may see that
$$\psi(t^{1-n}\xi_1, t \xi_2, t\xi_3, \dots ,t\xi_n) = \phi(\xi_1,\dots,\xi_n) e^{i A(t)}$$
for any $\xi_1,\dots,\xi_n,$ since we may normalize the $\xi_i$:s with
$(\xi_1 \xi_2 \cdots \xi_n)^{1/n}$ and move this to the other side ($\phi$).
Now, changing $t$ will not change the modulus of the product of the parameters to the function, so it may only affect the argument, and hence the form above.

$$\psi - n \xi_1 \psi'_1 = \psi i A'(1) $$
Now, this is a differential equation, which is easy to solve,
one sees that $\psi = \xi_1^\beta F(\xi_2,\dots,\xi_n)$ for some constant $\beta$ and unknown function $F$. However, this argument can be made for any of the variables,
yielding the desired result.

We do need that $\psi$ is differentiable, but this should be true almost everywhere for algebraic functions.

An entire algebraic function is just a polynomial function. So you know that it's a sum of terms of that type. (Unless it might have poles, in which case it's a rational function.)

Fix the terms $\xi_1,...,\xi_{n-1}$ at roots of unity and let $xi_n$ vary. Then $\xi_n$ is a polynomial that takes roots of unity. Therefore it must be a constant power of $\xi_n$ times a root of unity. (by Schwartz reflection) By continuity, the power cannot depend on $\xi_1,...,\xi_{n-1}$, so the only terms must have $\xi_n$ that power. We can continue this argument for each $\xi_i$, thus proving that the expresion consists of only a single term.

So, up to multiplication by a root of unity, it must be the option you gave.

If your function is allowed to have poles, this is false. Let $f$ be any rational linear transformation of the Riemann sphere that preserves the roots of unity, and let $g$ be a homogeneous polynomial of degree zero, then multiplying by $f\circ g$ will preserve all the properties given, and you can make your function arbitrarily complicated like this.
–
Will SawinFeb 2 '12 at 18:20

well maybe not arbitrarily. But extremely.
–
Will SawinFeb 2 '12 at 18:20

@Will Sawin: A polynomial of degree zero? That sounds very much like a constant...
–
Per AlexanderssonFeb 2 '12 at 22:09

oh sorry. i meant a rational function of total degree zero like $\zeta_1^2\zeta_2^{-1}\zeta_3^{-1}$
–
Will SawinFeb 3 '12 at 7:22

@Will Sawin: But the rational linear transform must be homogeneous as well...
–
Per AlexanderssonFeb 4 '12 at 11:24